- How to Find the Equation of a Circle
How to find the equation of a circle is an important concept in the realm of geometry. Embarking on exploring the elegance of geometry, this article will delve into the details of the circle. Circles are everywhere, from the celestial bodies in the sky to the wheels on which our cars run, making understanding their mathematical representation indispensable.
In this article, we’ll explore the methods and strategies for deriving the equation of a circle, a powerful tool in both pure and applied mathematics.
From simple geometric relations to complex applications, we’ll illustrate how the coordinates of the center and the length of the radius can define a circle’s equation. Whether you’re a mathematics enthusiast, a curious student, or an educator seeking clarity, we invite you on this intriguing journey into the world of circular reasoning.
How to Find the Equation of a Circle
To find the equation of a circle with center and radius , we use the following formula:
(x – h)² + (y – k)² = r²
- (h, k) is the center of the circle.
- r is the radius of the circle.
To find the equation of a circle, you need to know the center and the radius. If you know the coordinates of the center (h, k) and the radius (r), you substitute these values into the equation.
However, if you are given different information, such as the coordinates of points on the circle, you may need to use these points first to determine the center and radius. For instance, if you are given three points on the circle, you can use them to find the equation of the circle through methods involving distances and perpendicular bisectors.
Below we present a generic representation of the circle in figure-1.
In another case, if the circle equation is given in the general form Ax² + By² + Cx + Dy + E = 0, you may need to complete the square to transform it to the standard form.
Remember that, in the context of the equation, x, and y represent any point on the circle, h and k represent the circle’s center, and r represents the radius. This equation encapsulates the definition of a circle as the set of all points a fixed distance (the radius) from a given point (the center).
The equation of a circle is fundamental to understanding its properties. The equation itself is based on the definition of a circle: a set of points that are equidistant (the radius) from a fixed point (the center).
Let’s explore the properties of the circle and how they relate to its equation:
The center of the circle is given by the point (h, k) in the standard equation of a circle, (x – h)² + (y – k)² = r². The coordinates h and k can be any real numbers. The center point can be found directly from the equation in this standard form.
The value r in the standard equation gives the circle’s radius. It is the constant distance from the center to any point on the circle. Like the center, the radius can be found directly from the standard equation of a circle. Note that the radius must be a positive real number.
Points on the Circle
Any point (x, y) that satisfies the equation (x – h)² + (y – k)² = r² lies on the circle. These points can be found by substituting x or y values into the equation and solving for the corresponding y or x values.
Completing the Square
If a circle equation is given in the general form, Ax² + By² + Cx + Dy + E = 0, it can be converted into standard form by a process known as completing the square. This process rearranges and simplifies the equation to identify the center (h, k) and the radius r.
Diameter, Circumference, and Area
While these properties are not directly visible from the equation, they can be calculated using the radius, which is part of the equation. The diameter is twice the radius, the circumference is 2πr, and the area is πr².
Remember, the equation of a circle provides a roadmap to understanding the circle’s properties. It is a crucial tool in geometry and algebra for describing and investigating the nature of circles.
The ability to find the equation of a circle has a wide range of applications across numerous fields. Here are some examples:
Physics and Engineering
Circles describe the motion of objects in circular paths or orbits, such as planets, electrons around a nucleus, or objects in rotational motion. Engineers use circle equations in designing circular objects or paths, such as wheels, gears, and roundabouts.
Computer Graphics and Game Design
The equation of a circle is used to create round objects and effects or to calculate distances and collisions in games. Algorithms like the Midpoint Circle Algorithm use the equation of a circle to draw circular paths on the pixel grid of a screen.
Geography and GPS Technology
The concept of ‘circles of latitude’ describes the division of the Earth. In GPS technology, the equation of a circle (or sphere, in three dimensions) is used in trilateration to calculate a user’s location from the signals of multiple satellites.
Mathematics and Education
The equation of a circle is indeed a fundamental concept in geometry, algebra, and trigonometry. It is a basis for understanding and applying various mathematical concepts, including the Pythagorean theorem, functions, and complex numbers. By exploring the equation of a circle, students can develop a deeper understanding of these mathematical principles and their interconnectedness.
The orbits of celestial bodies are often approximated as circles (or ellipses, which are related). For instance, the transit method for detecting exoplanets involves observing the dip in a star’s brightness as a planet transits in front of it, which relies on understanding the planet’s circular path.
Architecture and Design
Circles are widely used in design due to their aesthetic appeal and symmetry. The ability to calculate the equation of a circle can help in creating accurate designs and models.
For a circle with a center at (2, -3) and a radius of 4, find the equation of the circle.
Substitute h = 2, k = -3, and r = 4 into the standard equation:
(x – 2)² + (y + 3)² = 4²
(x – 2)² + (y + 3)² = 16
Compute the equation of a circle with a center at the origin (0,0) and a radius of 5.
Substitute h = 0, k = 0, and r = 5 into the standard equation:
(x – 0)² + (y – 0)² = 5²
x² + y² = 25
Compute the equation of a circle with a center at (-1,2) and a point on the circle at (2,4).
First, find the radius using the distance formula between the center and the given point:
r = √[(2 – (-1))² + (4 – 2)²]
r = √
r = 3
Then substitute h = -1, k = 2, and r = 3 into the standard equation:
(x + 1)² + (y – 2)² = 3²
(x + 1)² + (y – 2)² = 9
Compute the equation of a circle passing through the origin (0,0) and having the center at (0, 4).
The radius is the distance from the center to a point on the circle (the origin):
r = √[(0 – 0)² + (0 – 4)²]
r = √
r = 4
Substitute h = 0, k = 4, and r = 4 into the standard equation:
x – 0)² + (y – 4)² = 4²
x² + (y – 4)² = 16
Given the equation, x² + y² – 6x + 8y – 9 = 0, convert it to the standard form of a circle and find the center and radius.
We can reorganize and complete the square:
x² – 6x + y² + 8y = 9
(x – 3)² – 9 + (y + 4)² – 16 = 9
(x – 3)² + (y + 4)² = 36
So, the center is at (3, -4), and the radius is √36 = 6.
Compute the equation of a circle with diameter endpoints at (2, 4) and (6, 8).
First, find the center by taking the midpoint of the endpoints:
h = (2 + 6)/2
h = 4
k = (4 + 8)/2
k = 6
Then, find the radius, which is half the length of the diameter:
r = √[(6 – 2)² + (8 – 4)²]/2
r = √
r = 4
Substitute h = 4, k = 6, and r = 4 into the standard equation:
(x – 4)² + (y – 6)² = 4²
(x – 4)² + (y – 6)² = 16
Compute the equation of a circle that touches the x-axis at the origin (0,0) and passes through the point (1,1).
Since the circle touches the x-axis at the origin, the center must be of the form (0, r). The radius r is the distance from the center to the point on the circle (1,1):
r = √[(1 – 0)² + (1 – r)²]
Solving the equation r² = 1 + 1 – 2r gives:
r = 1
Substitute h = 0, k = 1, and r = 1 into the standard equation:
(x – 0)² + (y – 1)² = 1²
x² + (y – 1)² = 1
Given the equation, 2x² + 2y² – 8x + 6y – 1 = 0, convert it to the standard form of a circle and find the center and radius.
Divide through by 2 and reorganize to complete the square:
x² – 4x + y² + 3y
= 0.5 (x – 2)² – 4 + (y + 1.5)² – 2.25
= 0.5 (x – 2)² + (y + 1.5)²
So, the center is at (2, -1.5), and the radius is √5.75 ≈ 2.4.
All images were created with GeoGebra.